Say we have a mixture of 10 mL of methane and propane. For a complete combustion, we need 41 mL of O2. Calculate the composition of the mixture in terms of volume, if atmospheric temperature and pressure are constants. The equations are: CH4 + O2 --> CO2 + H2O C3H8 + O2 --> CO2 + H2O I do not understand the question: am I asked to calculate a percentage of volumes, e.g. VCH4/VC3H8, or simply to find the volumes of CH4 and C3H8 separately (if so, how?)? Am I considering the reactives or the products of the reaction? These calculations must come in somewhere in the solution: I know that all gazes have the same molecular volume at normal temperature and pressure: 22.4L/mol. Therefore I can determine the quantity of moles of each component. In the initial mixture (before reaction), there is: 0.224mol of CH4 + C3H8 0.918mol of O2. Thx. for helping!